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In deriving a general propagator to the time-dependent ($H = H(t)$) Hamiltonian problem, Shankar works to first order in $\Delta = T/N$ (a small time interval for large $N$) and argues that by integrating the Schrodinger Equation over the interval $\Delta$ we get:

$$|\psi(\Delta)\rangle \approx |\psi(0)\rangle + \Delta \frac{d}{dt}|\psi(0)\bigg|_{t=0} = |\psi(0)\rangle - \frac{i\Delta}{\hbar}H(0)|\psi(0)\rangle = \left(1-\frac{i\Delta}{\hbar}H(0)\right)|\psi(0)\rangle \approx e^{-i\Delta H(0)/\hbar}|\psi(0)\rangle.$$

Then by integrating over each individual interval of width $\Delta = T/N$ gives:

$$|psi(t)\rangle = \prod_{n=0}^{N-1}\exp[-\frac{i\Delta}{\hbar}H(n\Delta)]|\psi(0)\rangle.$$

So far, I'm on board. However, he says that we can't simply turn the product into a sum in the exponent and take a limit as $N \rightarrow \infty$, get a nice integral representation and call it day. He says we can't do this $[H(t_1), H(t_2)] \neq 0$ so by Baker-Campbell-Hausdorff there will be additional terms. However, in the limit as $N \rightarrow \infty$ and so $\Delta \rightarrow 0$, can we not say that each $H(n\Delta)$ will commute with its neighbor $H((n+1)\Delta)$, and so argue that the integral representation is valid?

Qmechanic
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1 Answers1

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Let's call the evolution operator $U(t_2, t_1)$ for convenience.

It is true that when $\Delta \rightarrow 0$, $$ U(t + \Delta, t) \sim e^{- i \hbar^{-1} \Delta H}. $$

It is also true, like you noted, that $$ \lim_{\Delta \rightarrow 0} \left[ U(t + 2 \Delta, t + \Delta); U(t + \Delta, t) \right] = 0. $$

However, there will be terms proportional to $\Delta$ in the commutator: $$ \left[ U(t + 2 \Delta, t + \Delta); U(t + \Delta, t) \right] = A(t) \Delta + \mathcal{O}(\Delta^2). $$

Now imagine "collapsing" the string of exponentials using BCH. You will pick up $$ e^{A(t) \Delta} $$ every time you collapse two neighboring exponentials into one. You will have $N$ such terms, all of them with a different $A(t)$.

The product of these terms is (with further corrections coming from BCH, which are $\mathcal{O}(\Delta^2)$) $$ e^{\sum A(t) \Delta} \sim e^{A (t_2 - t_1)}, $$ where I assumed that all $A(t)$ are of the same order. This does not go to $1$ in the $\Delta \rightarrow 0$ limit.

Now this may be an illustration of a fallacy in your argument, but it is by no means a proof of anything. The actual proof comes from comparing the answer you would get if you assumed noncommutativity doesn't influence the outcome to the right answer. The right answer is

$$ U(t_2, t_1) = \mathcal{T} \exp \left( -i \hbar^{-1} \intop_{t_1}^{t_2} H(t) dt \right), $$

while you would get a similar formula with an ordinary exponential instead of the time-ordered one. These are demonstrably different operators.

Prof. Legolasov
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